# Sketching the image of the function $\hat{h}.$

I am trying to solve a puzzle through solving a series of questions, here is the first version of the puzzle:

Is it possible to find 2 connected subsets $$A$$ and $$B$$ inside the square $$I \times I$$ such that $$(0,0),(1,1) \in A$$ and $$(0,1),(1,0) \in B$$ and $$A \cap B = \emptyset.$$

We know that the answer is YES. But we are guessing that the answer is NO if we required $$A$$ and $$B$$ to be path connected.Our intuition comes from the attitude of the Topologist's sine curve. Now we are trying to write a proof of this guess. During this path we wanted to prove the Jordan Curve Theorem. So I am trying to do so.

Here are the questions that I should have solved so far:

Proving the equivalence of the Criss - Cross theorem statement(ordinary one) to another statement.

Proving the criss cross theorem (generalized version). (I did not solve this question yet, so any help with the solution will be appreciated )

Also I know the proof of the ordinary version of the criss-cross theorem. Here is it:

Assume towards contradiction that we have such paths $$\alpha$$ and $$\beta$$ such that $$\alpha (s) \neq \beta (t)$$ for all $$s,t \in I.$$ This can be written as $$\alpha (s)- \beta (t) \neq 0$$ for all $$s,t \in I.$$ Which can further be rephrased by defining $$H: I \times I \rightarrow \mathbb{R}^2$$ by the formula $$H(s,t) = \alpha(s) - \beta (t).$$ And our assumption means that the function $$H$$ never takes the value $$0 \in \mathbb{R}^2;$$ in diagram language, we assume that there is a continuous function $$\hat{H}$$ making the following diagram commutative:

$$\require{AMScd} \begin{CD} I \times I @>{\hat{H}}>> \mathbb{R}^2 - \{0\}\\ @VVV @VVV \\ I\times I @>{H}>> \mathbb{R}^2 \end{CD}$$

where the arrow below $$\hat{H}$$ should be a dotted arrow because we are searching for this function. And I am not skillful in drawing commutative diagrams this is why I draw $$I \times I$$ 2 times because I do not know how to draw one dotted arrow coming out of $$I \times I$$ going directly to $$\mathbb{R} - \{0\}$$ my bad. Then my job is to show that there can be no such function $$\hat{H}.$$

Now, I want to solve this problem:

Suppose there were such a function $$\hat{H},$$ and write $$\hat{h}$$ for the restriction of $$\hat{H}$$ to the boundary of $$I \times I.$$ Sketch what the image of $$\hat{h}$$ might look like - remember that there is no guarantee or assumption that either $$\hat{H}$$ or $$\hat{h}$$ is injective, so put some possible self-crossing in your sketch.

I do not know how to sketch it. could anyone help me in doing so please?

The key observation is that you know a little bit about what $$\hat{h}$$ looks like.

Assume $$\alpha : (0,0)\to (1,1)$$ and $$\beta : (0,1)\to (1,0)$$ are your paths.

Then $$\hat{h}(0,0) = \alpha(0)-\beta(0) = (0,0)-(0,1) = (0,-1).$$ as $$t$$ runs from $$0$$ to $$1$$, we get $$\hat{h}(t,0) = \alpha(t)-\beta(0) = (q,r) - (0,1) = (q,r-1).$$ We know almost nothing about this path except that $$0\le q\le 1$$ and $$-1\le r\le 0$$ and $$(q,r)\ne (0,1)$$, so drawing any path that stays inside this square and doesn't touch the origin is acceptable.

Then $$\hat{h}(1,0) = (1,1)-(0,1) = (1,0)$$. For $$0\le s\le 1$$, $$\hat{h}(1,s)$$ stays in $$[0,1]\times[0,1]\setminus\{(0,0)\}.$$ Again, drawing any path that stays in here is acceptable. Next $$\hat{h}(1,1) = (1,1)-(1,0) = (0,1)$$, and as $$t$$ goes from $$1$$ back to $$0$$ on the next segment, $$\hat{h}(t,1)$$ stays inside $$[-1,0]\times [0,1]$$ and doesn't touch the origin.

Finally, $$\hat{h}(0,1) = (-1,0)$$, and as $$s$$ returns to $$0$$, $$\hat{h}(0,s)$$ stays inside $$[-1,0]\times[-1,0]$$ and doesn't touch the origin before returning to $$(0,-1)$$.

You'll notice that any path you try to draw that satisfies these requirements forms a counterclockwise loop about the origin that is nonzero in the fundamental group, contradicting the assumption of the existence of the extension $$\hat{H}$$.

Note that this proves the sets in your puzzle cannot exist if they are path connected. Not sure why you would need the Jordan curve theorem.

• Maybe at the end of my question # 16 I think it will be clear why we would need Jordan curve theorem. – user778657 Apr 27 '20 at 2:23
• what do you mean by "nonzero in the fundamental group"? and how this will contradict the assumption of the existence of the extension $\hat{H}$? – user778657 Apr 28 '20 at 2:46
• @Smart20 I mean not contractible in the ambient space, and if $\hat{H}$ existed, it would give a contraction of the loop. – jgon Apr 28 '20 at 3:01